Prove that a positive monic polynomial with even degree has a minima but not a maxima.

Question

Prove that a positive monic polynomial with even degree has a minima but not a maxima. You may use the fact that there are finitely many critical points.

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Let's take the worst case where the first derivative has a single zero.

Let the polynomial be $f(x)$

We know $f'(c) = 0$ because $f'(x)$ is a odd degree polynomial.

Therefore we have a interior critical point.

Also by $\lim_{x \to \pm\infty} f(x)= + \infty$ we get no maxima.

Therefore $c$ is either minima or nothing.

Out of first/second derivative and direct test, I don't know what to apply here.

I thought about EVT but I don't know $[a,b]$ here.

What should I do now ?

Answer 1

Since $f$ is even degree, as you pointed out one has $\lim_{x\to-\infty} f(x)=\infty$ in addition to $\lim_{x\to\infty} f(x)=\infty$. Now choose $a$ small enough and $b$ big enough so that so that $f(a)>f(c)$ and $f(b)>f(c)$ where $c$ is the point you found. Then apply the extreme value theorem on the interval $[a,b]$ to get a local minimum.

If in addition one wants to find a global minimum, choose $a,b$ so that $f$ is decreasing on $(-\infty,a)$ and increasing on $(b,\infty)$.

Answer 2

Hint: Let $a = f(0)$. Since $\lim_{x\to \pm \infty} = \infty$, there exists $M > 0$ such that $$\vert x \vert > M \implies \vert f(x) \vert > a.$$ Now apply the extreme value theorem.

Answer 3

For both large positive and large negative values the leading term dominates (and turns the polynomial positive). Thus it has no maximum. But then it also must have a minimum because the polynomial cannot be unbounded over a small interval (lest it have an asymptote, which polynomials don't).